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I would like to know if there is a risk of damaging a 4AWG Battery wire (worst case fire hazard) which is rated for 60V, 150ADC if 300Amps @5VDC is passed for only 2 seconds? I am working on a test solution that will supply the UUT with 300 Amps in the production environment and current will be routed through the cables for 2 seconds max, at least 10 times per UUT test run. The situation is similar to starting a car as the starter motor takes a lot of current from the car battery.

The Length of the battery wires is going to be less than 10ft. Each unit is expected to take 5 minutes however, the high current application is not expected to last more than 15 seconds overall. The current will be applied as a step response for a maximum of 2 seconds. The test will be conducted at ambient temperature. The cable has polyvinylchloride (PVC) type insulation.

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  • \$\begingroup\$ Difficult to answer as you haven't specified the wire insulation, length of wire, ambient conditions and how long between tests (how long does the wire have to cool down between tests?). All of these may affect whether the wire will be able to handle a current twice its rating. \$\endgroup\$
    – Barry
    Oct 25, 2020 at 22:01
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    \$\begingroup\$ 2 minutes then yes. 0.2 seconds then no. 2 seconds is probably below the thermal time constant but not by a huge margin. \$\endgroup\$
    – winny
    Oct 25, 2020 at 22:09
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    \$\begingroup\$ You are exceeding the manufacturer's maximum ratings for the wire by a factor of 2. You must assume that the wire can be damaged by this. People here can give you opinions about whether it is likely or unlikely, based on their experience, but at the end of the day you have to respect the manufacturer's limits. \$\endgroup\$ Oct 25, 2020 at 22:26

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In practice, the insulation won't get hot enough to melt.

You can get an estimate of how long the cable can withstand a current by applying the adiabatic equation.

Given that 4AWG is around 25mm², and PVC cable is normally rated for 70°C. Assume a conservative value of k (from the web site linked above) of 103 for PVC copper cable that's already hot.

We get t = k²A² / I² = 103².25² / 300² = 73 seconds. Which is well above 2 seconds!

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  • \$\begingroup\$ Thank you @Simon B. Very helpful resource. \$\endgroup\$
    – Rav
    Oct 26, 2020 at 1:46
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If we assume that the 2 second pulses are close enough together that no significant cooling occurs between pulses, then we can calculate the temperature rise due to the energy added to the wire.

Assume 1 foot of wire. The length doesn't matter as long as all the values are adjusted for the length.

enter image description here

A temperature rise of 21 degC should be acceptable.

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  • \$\begingroup\$ Thank you very much @Mattman994. \$\endgroup\$
    – Rav
    Oct 26, 2020 at 1:46

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